Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 304. (Read 240606 times)
Hi, I still have to understand, how do they use that script in python with the gpu.
80 to 39vbryGKfcpuch6RLWa97BqG3rKFQQWJh7 and stop
85 to 3Qpip6a2oYgbrcomVpeaUZ4PVXsGVshuec and stop
so who found #80 and #85? no one has taken credits or posted their private keys here yet 85 to 3Qpip6a2oYgbrcomVpeaUZ4PVXsGVshuec and stop
are there other local (russian etc) threads about this topic?
An interesting situation. The numbers 65, 70, 75, 80, 85 were transferred to different multi-addresses 3 ...., and no further movement. If there was one person. That would merge on one address, and probably would use these funds. And so some strange movement.
I think they were claimed by different person #65 by arulbero
#70 by pikachunakapika
#75 by pikachunakapika
I'm just guessing here by reading past posts in this thread
An interesting situation. The numbers 65, 70, 75, 80, 85 were transferred to different multi-addresses 3 ...., and no further movement. If there was one person. That would merge on one address, and probably would use these funds. And so some strange movement.
65 to 3GVSoQar7WnVMFjchpkNkh9UwkJbbdeCwM and stop
70 to 3MpeFYTuwfrxSnUXNmTnY3XTgvhPp5k7vF and stop
75 to 3HrUTaDiyKaGMyVycK8fynRGpAPV64AtV7 and stop
80 to 39vbryGKfcpuch6RLWa97BqG3rKFQQWJh7 and stop
85 to 3Qpip6a2oYgbrcomVpeaUZ4PVXsGVshuec and stop
In BCH
65 to qrvyzatwha3v3rjjtrx52qv8fe9kppcxcsfj2l6u7x and stop
70 to qqhqyz2lk0vg54x496cluhsxllz5kzjx45uy2lk8nj and stop
75 to qpwpa9twauk93ytd9rk6cgnutmgag6jldqfe7rkqy0 and stop
80 to qrwze4p5tg70jm00vvea797czsyhmw87qvwezya0ys and to qpzhkgrq5evgpsqczah9u32f7qx4ggw2ru2wuxj2ra and stop
85 to qz9tf4gmk869jv959lnyuhx5hevuj0kdxq6u8zrccs and stop
65 to 3GVSoQar7WnVMFjchpkNkh9UwkJbbdeCwM and stop
70 to 3MpeFYTuwfrxSnUXNmTnY3XTgvhPp5k7vF and stop
75 to 3HrUTaDiyKaGMyVycK8fynRGpAPV64AtV7 and stop
80 to 39vbryGKfcpuch6RLWa97BqG3rKFQQWJh7 and stop
85 to 3Qpip6a2oYgbrcomVpeaUZ4PVXsGVshuec and stop
In BCH
65 to qrvyzatwha3v3rjjtrx52qv8fe9kppcxcsfj2l6u7x and stop
70 to qqhqyz2lk0vg54x496cluhsxllz5kzjx45uy2lk8nj and stop
75 to qpwpa9twauk93ytd9rk6cgnutmgag6jldqfe7rkqy0 and stop
80 to qrwze4p5tg70jm00vvea797czsyhmw87qvwezya0ys and to qpzhkgrq5evgpsqczah9u32f7qx4ggw2ru2wuxj2ra and stop
85 to qz9tf4gmk869jv959lnyuhx5hevuj0kdxq6u8zrccs and stop
solved #85
Congrats to the solver. I don't have the hardware to get those priv keys
Congrats to the solver. I don't have the hardware to get those priv keys
...
Calculate the time to restore the keys of the puzzle.
The use of the program break_short(brsh) is supposed
1core brsh = 0,27 Mkeys/s
128cr brsh = 34,5 Mkeys/s
Time is indicated for calculation only for one table(BS/GS), but they should be considered two, that is, double the actual time.
Ideal calculations suggest that you have enough RAM (no penalties)
....
Not one) #70 success calc.
0290E6900A58D33393BC1097B5AED31F2E4E7CBD3E5466AF958665BC0121248483
00000000000000000000000000000000000000000000003*9*4*6*3*A*C*E*1
Our legendary master, why so long, maybe it's time to clean up the Xeon registers?)
It took almost 12 hours to find the #70 on my system (Intel® Xeon® CPU E3-1505M v6 @ 3.00GHz + 32 GB RAM). No GPU, only CPU (1 core, my program is not multithread).
Only 6 minutes to retrieve the #65.
pikachunakapika
there is 1080TI 4
PM
there is 1080TI 4
PM
#75: 4c5ce114686a1336e07
I have a very slow GPU (965m). For 80+ I need better GPU.
I have a very slow GPU (965m). For 80+ I need better GPU.
How much cpu time for case #75 and what are your previsions for case #80 and #85?
If you can make a AMD opencl version, I have a lot of hardware for that (20 Rx480 ready)
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#75: 4c5ce114686a1336e07
I have a very slow GPU (965m). For 80+ I need better GPU.
I have a very slow GPU (965m). For 80+ I need better GPU.
Thank you very much. Now it seems that everything should be ok.
As for the RAM demand ... - no one orders to run it from the beginning, since I am interested in the scope, eg 80 - I can skip 79 previous ones, freeing up memory by the whole 2 ^ 79
As for the RAM demand ... - no one orders to run it from the beginning, since I am interested in the scope, eg 80 - I can skip 79 previous ones, freeing up memory by the whole 2 ^ 79
How to modify the current Baby-step-giant-step code to be able to compile on the current keyspace? I realize how long it will last and how much RAM will need.
with these declarations:
typedef struct hashtable_entry {
uint128_t x;
uint128_t exponent;
} hashtable_entry;
#define HASH_SIZE (2*GSTEP)
hashtable_entry table[HASH_SIZE];
using
#define GSTEP ((uint128_t)1<<32)
requires 256 Gb RAM !!!
to solve case #85, I assume that you need #define GSTEP ((uint128_t)1<<42)
that's 256*2*2*2*2*2*2*2*2*2*2 = 256 Tb RAM
Doesn't work.
warning: format ‘%lx’ expects argument of type ‘long unsigned int’, but argument 4 has type ‘uint128_t {aka __int128 unsigned}’ [-Wformat=]
printf("Found private key %2d: %16lx or %16lx\n", j + 1,
Easy to solve. Just rewrite the print section by separating lo and hi 64bits.
this is what I did
Code:
while (table[entry].exponent != 0) {
if (table[entry].x == (uint64_t) xst.n[2]) {
uint128_t key = (uint128_t) i * (uint128_t) (2 * GSTEP);
uint128_t key1 = key - table[entry].exponent ;
uint128_t key2 = key + table[entry].exponent;
uint64_t key1lo = key1;
uint64_t key1hi = (key1 >> 64);
uint64_t key2lo = key2;
uint64_t key2hi = (key2 >> 64);
printf("Found private key %2d: %lx %lx or %lx %lx\n", j + 1, key1hi,key1lo,key2hi,key2lo);
next++;
..................................
How to modify the current Baby-step-giant-step code to be able to compile on the current keyspace? I realize how long it will last and how much RAM will need.
with these declarations:
typedef struct hashtable_entry {
uint128_t x;
uint128_t exponent;
} hashtable_entry;
#define HASH_SIZE (2*GSTEP)
hashtable_entry table[HASH_SIZE];
using
#define GSTEP ((uint128_t)1<<32)
requires 256 Gb RAM !!!
to solve case #85, I assume that you need #define GSTEP ((uint128_t)1<<42)
that's 256*2*2*2*2*2*2*2*2*2*2 = 256 Tb RAM
Doesn't work.
warning: format ‘%lx’ expects argument of type ‘long unsigned int’, but argument 4 has type ‘uint128_t {aka __int128 unsigned}’ [-Wformat=]
printf("Found private key %2d: %16lx or %16lx\n", j + 1,
pikachunakapika
hello give try your program test your program thanks
hello give try your program test your program thanks
#70: 349b84b6431a6c4ef1
Time for calculations!
hoho, love it
==================
1)We calc 1st pubkey table and save in RAM.
2)We calc 2st pubkey table and on the fly compare it with 1st table in ram without save.
3)If a match is found, the privkey is ready.
The size of the tables is the same and less than the square root of the bit size of the key being flushed.
The comparison procedure in RAM is essentially the multiplication of two square roots, example 2^40 * 2^40 = 2^80.
2^80 is very difficult to calculate, but 2^40 times easier. Ok?
RAM is necessary because the comparison should be as FAST as possible.
Computational complexity
When there is enough RAM, each puzzle requires x2 more calculations.
But in case of a lack of ram, each puzzle the computational complexity increases x4 (exponential growth).
If there is not enough ram, you can compare BSvsGS by calculating BS/GS in chunks.
But the price is high - an exponential increase in computing.
For comparison with the next piece of BS loaded into memory, we are forced to re-calculate the GS completely.
An illustrative example showing dimensions.
Increase computation that you have 16Gb free RAM (penalties if not enough)
norm if enough ram
x1->x2->x4->x8->x16
penalties if not enough ram
x1->x3->x10->x36->x136
Due to this fact, the GPU calculations will be much less efficient.
The width of the memory bus is crucial, as the most powerful data exchange (Vega/Volta) is assumed.
==================
Calculate the time to restore the keys of the puzzle.
The use of the program break_short(brsh) is supposed
1core brsh = 0,27 Mkeys/s
128cr brsh = 34,5 Mkeys/s
Time is indicated for calculation only for one table(BS/GS), but they should be considered two, that is, double the actual time.
Ideal calculations suggest that you have enough RAM (no penalties)
* On Linux operating systems, performance is 25% higher than on Windows.
* Assumes server instance AmazonAWS 4Tb ddr4 128cores (x1e.32xlarge)
* The actual performance of 128 cores can be reduced to 96 cores if the server has 64 real cores of 2 logical cores each.
==================
Calculations suggest that you have 4Tb of memory (penalties if not enough)
* penalty calculation formula
penalty = (((norm * norm) + norm) / 2)
Now restore the chronology of events
May 31, the Author deliberately discloses the pubkeys in every fifth transaction in order to update and improve the puzzle.
On June 7, a vigilant certain Mr.X gets #65, then #70, #75, #80 within 4 days.
By the above calculation it is easy to understand how he did it.
He modified break_short, adding 64-128bit, multi-core and the ability to calculate BS in chunks.
(even a novice C programmer can easily add these modifications to the program)
Then he rented amazonaws x1e.32xlarge with 2-4Tb for 4 days (250$/day), spending 1k $ and earned 20k $
I hate you, but you're well done)
How is it going in London?
You know, Bivonas is not the best place)
Why didn't he find #85? As can be seen from the table it is unprofitable! (without significant improvement of Software or Server)
Arulbero has already wrote some posts of improving the Software, let's leave it.
I think that beyond #85 it will be very difficult to recover the private key, even with 1 TB of RAM (with the Baby-Giant Step algorithm).
Nvme ssd as a swap?
SSD is much slower than RAM. (DDR4 47GB/s)
Let's talk about Servers.
We are not so important processor or videocard. But the size of RAM is very important to us.
The problem is that more than 4Tb RAM is not mass produced.
Consider a few ideas around this.
NVMe has approximately the same bandwidth as DDR4 RAM, but the access speed is much lower.
A HDD has a response time of about 10ms, an SSD will respond in 0.1ms, an Optain will respond in 8micros, a RAM will respond in 50ns.
0.1ms is equal to 100,000 ns.
This means that RAM can serve data in memory 1000 times faster than NVMe disk.
I checked the paging file(swap) on M.2 RAID0 NVMe - it's useless. its slooooooooow (100Kkey/s)
But there is news from Western Digital
Memory Extension Drive
Ultrastar DC ME200 1,2,4Tb
https://blog.westerndigital.com/how-ultrastar-memory-extension-drive-works/
https://www.westerndigital.com/products/data-center-drives/ultrastar-dc-me200-memory-extension-drive
According to WD, if we take the appeal to 786 GB of DRAM for pure performance, which is estimated at 1,080 million transactions per second, then a 3: 1 mix (576 GB of Ultrastar and 192 GB of DRAM) will drop to 91% of performance or to 983 thousand transactions, and at a ratio of 7: 1 (672 GB Ultrastar and 96 GB DRAM), the performance will decrease to 85% or 918 thousand transactions per second.
Intel Optain(Optane)
Optain looks like an acceleration cache for hard drives.
It can also work in the expansion mode of RAM (!).
It will be useful for a large number of small file accesses(!).
Optain has less capacity than SSD.
A 7th generation Intel processor is required, so it does not improve performance for older PCs.
Optane can to 7 microseconds at readings.
Optane can only get QD1 performance. This is impressive (5 times faster than current generation SSM NVMe).
Intel Optane SSD 750Gb
https://itpeernetwork.intel.com/optane-intel-memory-drive-technology/
Optane DC Persistent Memory DIMM format
https://lenovopress.com/lp1066-intel-optane-dc-persistent-memory
0290E6900A58D33393BC1097B5AED31F2E4E7CBD3E5466AF958665BC0121248483
00000000000000000000000000000000000000000000003*9*4*6*3*A*C*E*1
Our legendary master, why so long, maybe it's time to clean up the Xeon registers?)
hoho, love it
==================
can you explain in short why the size of RAM matters in this algorithm?
The essence of the BSGS (baby-step-gigant-step) algorithm in simple words.1)We calc 1st pubkey table and save in RAM.
2)We calc 2st pubkey table and on the fly compare it with 1st table in ram without save.
3)If a match is found, the privkey is ready.
The size of the tables is the same and less than the square root of the bit size of the key being flushed.
The comparison procedure in RAM is essentially the multiplication of two square roots, example 2^40 * 2^40 = 2^80.
2^80 is very difficult to calculate, but 2^40 times easier. Ok?
RAM is necessary because the comparison should be as FAST as possible.
Computational complexity
When there is enough RAM, each puzzle requires x2 more calculations.
But in case of a lack of ram, each puzzle the computational complexity increases x4 (exponential growth).
If there is not enough ram, you can compare BSvsGS by calculating BS/GS in chunks.
But the price is high - an exponential increase in computing.
For comparison with the next piece of BS loaded into memory, we are forced to re-calculate the GS completely.
An illustrative example showing dimensions.
Increase computation that you have 16Gb free RAM (penalties if not enough)
Code:
#61,#62 16Gb/16=1 x1
[BS]
[GS]
.......................................
#63,#64 32Gb/16=2 x3 ((2*2+2)/2)
[BS][BS]
[GS][GS][GS][GS]
.......................................
#65,#66 64Gb/16=4 x10 ((4*4+4)/2)
[BS][BS]
[BS][BS]
[GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS]
.......................................
#67,#68 128Gb/16=8 x36 ((8*8+8)/2)
[BS][BS][BS][BS]
[BS][BS][BS][BS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
.......................................
#69,#70 256Gb/16=16 x136 ((16*16+16)/2)
[BS][BS][BS][BS]
[BS][BS][BS][BS]
[BS][BS][BS][BS]
[BS][BS][BS][BS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
[GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS][GS]
norm if enough ram
x1->x2->x4->x8->x16
penalties if not enough ram
x1->x3->x10->x36->x136
Due to this fact, the GPU calculations will be much less efficient.
The width of the memory bus is crucial, as the most powerful data exchange (Vega/Volta) is assumed.
==================
How long did it take and how much mem does it require ...
I realize how long it will last and how much RAM will need.
what equipment is needed to solve the puzzle and what software?
Calculate the time to restore the keys of the puzzle.
The use of the program break_short(brsh) is supposed
1core brsh = 0,27 Mkeys/s
128cr brsh = 34,5 Mkeys/s
Time is indicated for calculation only for one table(BS/GS), but they should be considered two, that is, double the actual time.
Ideal calculations suggest that you have enough RAM (no penalties)
Code:
2^x |#puzzle| memory | 1core brsh | 128cr brsh |
------------------------------------------------
2^23(#47,#48) 128(Mb) 30(sec) | 0() |
2^24(#49,#50) 256(Mb) 60(sec) | 0,5(sec)|
2^25(#51,#52) 512(Gb) 2(min) | 1(sec)|
2^26(#53,#54) 1(Gb) 4(min) | 2(sec)|
2^27(#55,#56) 2(Gb) 8(min) | 4(sec)|
2^28(#57,#58) 4(Gb) 16(min) | 8(sec)|
2^29(#59,#60) 8(Gb) 32(min) | 16(sec)|
2^30(#61,#62) 16(Gb) 1(h) | 32(sec)|
2^31(#63,#64) 32(Gb) 2(h) | 1(min)|
2^32(#65,#66) 64(Gb) 4(h) | 2(min)|
-----double-ram-x2(Gb)-because-need-x32->x64----
2^33(#67,#68) 256(Gb) 8(h) | 4(min)|
2^34(#69,#70) 512(Gb) 16(h) | 8(min)|
2^35(#71,#72) 1(Tb) 32(h) | 16(min)|
2^36(#73,#74) 2(Tb) 2,5(d) | 32(min)|
2^37(#75,#76) 4(Tb) 5(d) | 1(h) |
2^38(#77,#78) 8(Tb) 10(d) | 2(h) |
2^39(#79,#80) 16(Tb) 20(d) | 4(h) |
2^40(#81,#82) 32(Tb) 40(d) | 8(h) |
2^41(#83,#84) 64(Tb) 3(m) | 16(h) |
2^42(#85,#86) 128(Tb) 6(m) | 32(h) |
2^43(#87,#88) 256(Tb) 9(m) | 2,5(d) |
2^44(#89,#90) 512(Tb) 1,5(y) | 5(d) |
2^45(#91,#92 1(Pb) 3(y) | 10(d) |
2^46(#93,#94) 2(Pb) 6(y) | 20(d) |
2^47(#95,#96) 4(Pb) 12(y) | 40(d) |
2^47(#97,#98) 8(Pb) 24(y) | 3(m) |
2^47(#99,#100) 16(Pb) 48(y) | 6(m) |
* On Linux operating systems, performance is 25% higher than on Windows.
* Assumes server instance AmazonAWS 4Tb ddr4 128cores (x1e.32xlarge)
* The actual performance of 128 cores can be reduced to 96 cores if the server has 64 real cores of 2 logical cores each.
==================
Calculations suggest that you have 4Tb of memory (penalties if not enough)
Code:
2^x |#puzzle | memory | norm/penalty| 1core brsh | 128cr brsh |
----------------------------------------------------------------
2^37(#75,#76) 4(Tb)| 1/ 1| 5(d)/ 5(d)| 1(h)/ 1(h)|
2^38(#77,#78) 8(Tb)| 2/ 3| 10(d)/ 15(d)| 2(h)/ 3(h)|
----------------------------------------------------------------
2^39(#79,#80) 16(Tb)| 4/ 10| 20(d)/1,5(m)| 4(h)/ 10(h)|
2^40(#81,#82) 32(Tb)| 8/ 36| 40(d)/ 6(m)| 8(h)/ 36(h)|
2^41(#83,#84) 64(Tb)| 16/ 136| 3(m)/ 2(y)| 16(h)/ 6(d)|
----------------------------------------------------------------
2^42(#85,#86) 128(Tb)| 32/ 528| 6(m)/ 8(y)| 32(h)/ 22(d)|
2^43(#87,#88) 256(Tb)| 64/ 2080| 9(m)/ 24(y)|2,5(d)/ 3(m)|
----------------------------------------------------------------
2^44(#89,#90) 512(Tb)| 128/ 8256|1,5(y)/ 95(y)| 5(d)/ 10(m)|
2^45(#91,#92) 1(Pb)| 256/ 32896| 3(y)/385(y)| 10(d)/3,5(y)|
2^46(#93,#94) 2(Pb)| 512/ 131328| 6(y)/ ?(y)| 20(d)/ 14(y)|
----------------------------------------------------------------
2^47(#95,#96) 4(Pb)|1024/ 524800| 12(y)/ ?(y)| 40(d)/ 56(y)|
2^47(#97,#98) 8(Pb)|2048/2098176| 24(y)/ ?(y)| 3(m)/252(y)|
----------------------------------------------------------------
2^47(#99,#100) 16(Pb)|4096/8390656| 48(y)/ ?(y)| 6(m)/ ?(y)|
* penalty calculation formula
penalty = (((norm * norm) + norm) / 2)
How is this possible?
Now restore the chronology of events
May 31, the Author deliberately discloses the pubkeys in every fifth transaction in order to update and improve the puzzle.
On June 7, a vigilant certain Mr.X gets #65, then #70, #75, #80 within 4 days.
By the above calculation it is easy to understand how he did it.
He modified break_short, adding 64-128bit, multi-core and the ability to calculate BS in chunks.
(even a novice C programmer can easily add these modifications to the program)
Then he rented amazonaws x1e.32xlarge with 2-4Tb for 4 days (250$/day), spending 1k $ and earned 20k $
I hate you, but you're well done)
How is it going in London?
You know, Bivonas is not the best place)Why didn't he find #85? As can be seen from the table it is unprofitable! (without significant improvement of Software or Server)
Arulbero has already wrote some posts of improving the Software, let's leave it.
I think that beyond #85 it will be very difficult to recover the private key, even with 1 TB of RAM (with the Baby-Giant Step algorithm).
SSD is much slower than RAM. (DDR4 47GB/s)
Let's talk about Servers.
We are not so important processor or videocard. But the size of RAM is very important to us.
The problem is that more than 4Tb RAM is not mass produced.
Consider a few ideas around this.
NVMe has approximately the same bandwidth as DDR4 RAM, but the access speed is much lower.
A HDD has a response time of about 10ms, an SSD will respond in 0.1ms, an Optain will respond in 8micros, a RAM will respond in 50ns.
0.1ms is equal to 100,000 ns.
This means that RAM can serve data in memory 1000 times faster than NVMe disk.
I checked the paging file(swap) on M.2 RAID0 NVMe - it's useless. its slooooooooow (100Kkey/s)
But there is news from Western Digital
Memory Extension Drive
Ultrastar DC ME200 1,2,4Tb
https://blog.westerndigital.com/how-ultrastar-memory-extension-drive-works/
https://www.westerndigital.com/products/data-center-drives/ultrastar-dc-me200-memory-extension-drive
According to WD, if we take the appeal to 786 GB of DRAM for pure performance, which is estimated at 1,080 million transactions per second, then a 3: 1 mix (576 GB of Ultrastar and 192 GB of DRAM) will drop to 91% of performance or to 983 thousand transactions, and at a ratio of 7: 1 (672 GB Ultrastar and 96 GB DRAM), the performance will decrease to 85% or 918 thousand transactions per second.
Intel Optain(Optane)
Optain looks like an acceleration cache for hard drives.
It can also work in the expansion mode of RAM (!).
It will be useful for a large number of small file accesses(!).
Optain has less capacity than SSD.
A 7th generation Intel processor is required, so it does not improve performance for older PCs.
Optane can to 7 microseconds at readings.
Optane can only get QD1 performance. This is impressive (5 times faster than current generation SSM NVMe).
Intel Optane SSD 750Gb
https://itpeernetwork.intel.com/optane-intel-memory-drive-technology/
Optane DC Persistent Memory DIMM format
https://lenovopress.com/lp1066-intel-optane-dc-persistent-memory
You are the only one that has the working code for that
Not one) #70 success calc. 0290E6900A58D33393BC1097B5AED31F2E4E7CBD3E5466AF958665BC0121248483
00000000000000000000000000000000000000000000003*9*4*6*3*A*C*E*1
Our legendary master, why so long, maybe it's time to clean up the Xeon registers?)
How to modify the current Baby-step-giant-step code to be able to compile on the current keyspace? I realize how long it will last and how much RAM will need.
with these declarations:
typedef struct hashtable_entry {
uint128_t x;
uint128_t exponent;
} hashtable_entry;
#define HASH_SIZE (2*GSTEP)
hashtable_entry table[HASH_SIZE];
using
#define GSTEP ((uint128_t)1<<32)
requires 256 Gb RAM !!!
to solve case #85, I assume that you need #define GSTEP ((uint128_t)1<<42)
that's 256*2*2*2*2*2*2*2*2*2*2 = 256 Tb RAM
How to modify the current Baby-step-giant-step code to be able to compile on the current keyspace? I realize how long it will last and how much RAM will need.
Well, what is the sacramental meaning of the step through 5? And he just brute force? because, it seems, it is necessary that the address be translated to its public key lit up. Arulbero could not until there is no output, find the number or could?)).
The author of the puzzle did that transaction to spend from every fifth address.
My guess is that it is an experiment to see how far people can get when there is an available spend transaction versus addresses with no available spend transaction.
He has been known to visit this thread. Perhaps he will explain it.
Obviously it is much easier to get the private key when there is a spend transaction on the address. #1 through #61 took a long time whereas #65, #70, #75 and #80 were snatched up pretty soon after the author added the spend transaction to those addresses. I expect #85 will also be snatched up in due time.
As discussed #85, #90, #95, #100, #105, #110 are all within the realm of possibility given enough time and resources. It looks as if #115 would be a new world record so someone with enough equipment and motivation can probably get that one. Beyond that it is very iffy.
Note that since there is a spend transaction on these addresses we will probably see more of them before we ever see the next address without a spend transaction, #62. This is a HUGE difference in effort.
I think one of the take home messages here might be that due to this difference in effort and other factors having to do with privacy and the fungiblity of Bitcoin in general: do not reuse Bitcoin addresses. Bitcoin addresses should be used exactly twice: once to fund them and once to spend them - then never used again.
No not bitcrack. It does not work here.
Hi, i have an asus 1080ti
but do you have to use bitcrack?
but do you have to use bitcrack?
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